Find Value Using Vector Magnitude Relations

 

Question:

Let a and b be vectors of the same magnitude such that

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid }{\mid \mathbf{a}-\mathbf{b}\mid }=\sqrt{2}+1.$$

Then the value of

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid +\mid \mathbf{a}-\mathbf{b}\mid }{\mid \mathbf{a}+\mathbf{b}\mid -\mid \mathbf{a}-\mathbf{b}\mid }$$

is:

📷 Question Image:

Short Solution (Text):

Let

$$k=\frac{\mid \mathbf{a}+\mathbf{b}\mid }{\mid \mathbf{a}-\mathbf{b}\mid }=\sqrt{2}+1.$$

Divide numerator and denominator of the required expression by $|\mathbf{a}-\mathbf{b}|$:

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid +\mid \mathbf{a}-\mathbf{b}\mid }{\mid \mathbf{a}+\mathbf{b}\mid -\mid \mathbf{a}-\mathbf{b}\mid }=\frac{k+1}{k-1}\mathrm{.}$$

Substitute $k=\sqrt{2}+1$:

$$\frac{k+1}{k-1}=\frac{(\sqrt{2}+1)+1}{(\sqrt{2}+1)-1}=\frac{\sqrt{2}+2}{\sqrt{2}}=1+\frac{2}{\sqrt{2}}=1+\sqrt{2}\mathrm{.<span\; class="mord\; sqrt"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span><span\; class="mord">.</span>}$$

---

✅ Final Answer:

$$\boxed{1+\sqrt{2}}$$

---

📷 Solution Image: